I wanto know the smoothness of convex set in ${\bf R}^n$. 
  Recall the following definition.
  Definition : $X$ is a bounded closed convex set in ${\bf R}^n$ if for $x$, $y\in X$, the any $d$-minimizing geodesic from $x$ to $y$ lies in $X$ where $d$ is a distance function of $X$.

 That is, if $Y= S^{n-1}(1)$ and $X$ is convex then for $a$, $b\in X$, then $\frac{sa + (1-s)b}{|sa + (1-s)b |}$ is in $X$ for $0< s<1$ 

 Question 1) Does the boundary of $m$-dimensional bounded convex set has dimension $m-1$ ?


 Question 2) Is the following opinion is right ?

  $(\ast)$ My thought : Let $m\geq 2$. A $(m-1)$-dimensional boundary of a $m$-dimensional bounded closed convex set $X$ is smooth 
  except some $(m-2)$-dimensional set. 

   


  The motivation of this is as follows: In some paper, the Hausdorff measure of convex set in $S^{n-1}(1)$ is considered. 
   

  That is, in my thought convex set may be a set of noninteger Hausdorff dimension. 
  Am I right ? 

  If $\ast$ is right, then why does one consider the Hausdorff measure of convex set ? 

  Thank you in advance. 

[paper's content]-----------------------------------------------------

3.1 Proposition : $X$ is a closed convex set in $S^{n-1}(1)$ and $u$ is a point in $X$
Then area$ (X\cap {\bf H}_u) \geq \frac{1}{2} $area$ (X)$ where ${\bf H}_u = \{ p\in S^{n-1}(1) | p\cdot u \geq 0\}$

3.2 Note : If $X \subset S^{n-1}(1)$ is a convex $spherical$ set of Hausdorff dimension $d$, then $H^d(X\cap {\bf H}_u) \geq \frac{1}{2} H^d(X)$ where $H^d$ is the $d$-dimensional Hausdorff measure. 

Here there is the word "spherical".  I think that if we omit the word, then it is also fine.